ECE 421/521 Electric Energy Systems Power Systems Analysis I 2 Basic Principles. Instructor: Kai Sun Fall 2013

Transcription

1 ECE 41/51 Electric Energy Systems Power Systems Analysis I Basic Principles Instructor: Kai Sun Fall 013 1

2 Outline Power in a 1-phase AC circuit Complex power Balanced 3-phase circuit

3 Single Phase AC System it ( ) = I cos( ωt+ θ ) m i + vt ( ) = Vm cos( ωt+ θv) - Z Load 3

4 Phasor Representation vt ( ) = V cos( ωt+ θ ) = V cos( ωt+ θ ) m v v it ( ) = I cos( ωt+ θ ) = I cos( ωt+ θ ) m i i V= V θ v I= I θ i RMS phasors of v(t) and i(t) Reference A phasor is a complex number that carries the amplitude and phase angle information of a sinusoidal function Phasors represent sinusoidal signals of a common frequency (ω) as vectors in the complex plane w.r.t. a chosen reference signal. Phasor representation is a mapping from the time domain to complex number domain. 4

5 Instantaneous power delivered to the load: Using trigonometric identity 1 cos Acos B = cos( A- B) + cos( A+ B) θ = θ θ v V = V /, I = I / m [ ] i Impedance angle >0 for inductive load and <0 for capacitive load Root-mean-square (RMS) values m pt () = vtit ()() = VI cos( ωt+ θ )cos( ωt+ θ ) m m v i 1 = VmIm[ cos( θv θi) + cos( ωt + θv + θi) ] 1 = VmIm{ cos( θv θi) + cos[( ωt + θv) ( θv θi)] } 1 = VmIm{ cosθ + cos[( ωt + θv) θ] } 1 = VmIm[ cosθ + cos ( ωt + θv) cosθ + sin ( ωt + θv) sinθ ] p( t) = V I cos θ[1 + cos ( ωt + θv)] + V I sinθsin ( ωt + θv) p () t p () t R X 5

6 Example.1 Assume vt ( ) = 100cosωt Z = Ω Calculate the RMS current phasor: I = = o it ( ) = 80cos( ωt 60 ) 100 v(t)=v m cos ωt, i(t)=i m cos(ωt -60) 6000 p(t)=v(t) i(t) ωt, degree p R (t) Eq ωt, degree p X (t) Eq ωt, degree ωt, degree 6

7 p( t) = V I cos θ[1 + cos ( ωt + θv)] + V I sinθsin ( ωt + θv) p () t p () t R X Observations on p R (t): Oscillating at frequency *ω (twice of the source frequency) Changing between 0 and V I cosθ (always positive) Average value is V I cosθ It is the energy flow into the circuit Observations on p X (t): Oscillating at frequency *ω (twice of the source frequency) Changing between - V I sinθ and V I sinθ (either positive or negative) Average value is zero It is the energy borrowed & returned by the circuit. It does no useful work in the load but takes some line capacity 7

8 Real and Reactive Powers p( t) = V I cos θ[1 + cos ( ωt + θv)] + V I sinθsin ( ωt + θv) V I Apparent power def P = V I cosθ Real power or active power (average power) Unit ~ watt or W p () t p () t Unit ~ volt ampere or VA (kva or MVA) Power factor (PF): cosθ= cos(θ v -θ i ) R Lagging or leading when θ =θ v -θ i >0 or <0 X Q P def Q = V I sinθ Reactive power Unit ~ var (volt-ampere reactive). Some people use Var, VAR or VAr Q>0 or <0 when θ=θ v -θ i >0 or <0 8

9 Characteristics of instantaneous power p(t) p( t) = V I cos θ[1 + cos ( ωt+ θv)] + V I sinθsin ( ωt+ θv) For a pure resistor load (thermal load), θ=θ v -θ i =0, PF=1 (unity PF) P= V I, so all electric energy becomes thermal energy For a pure inductive load, θ=θ v -θ i =90, PF=0 P=0, so electric energy = constant (no transformation to other forms) p(t) oscillates between the source and the magnetic field associated with the inductive load For a pure capacitive load, θ=θ v -θ i =-90, PF=0 p () t p () t R = P[1 + cos ( ωt+ θ )] + Qsin ( ωt+ θ ) v P=0, so electric energy = constant (no transformation to other forms) p(t) oscillates between the source and the electric field associated with the capacitive load v X 9

10 Example.1 Z = Ω vt ( ) = 100cosωt o it ( ) = 80cos( ωt 60 ) RMS phasors: V = 0, I = 60 o o p( t) = V I cos θ[1 + cos ωt] + V I sinθsin ωt = 4000cos 60 [1 + cos ωt] sin 60 sin ωt () p () () () pr t X t pr t px t P Q 100 v(t)=v m cos ωt, i(t)=i m cos(ωt -60) 6000 p(t)=v(t) i(t) P ωt, degree 4000 p R (t) Eq ωt, degree 4000 p X (t) Eq P ωt, degree ωt, degree Borrowing Returning Q Q

11 What does absorbing or generating reactive power mean? Reactive power with a circuit element does not lead to real energy consumption Starting from the time of v(t)=v m, An inductive element (lagging PF) first absorbs and then returns the same amount of electric energy. Such a pattern is considered absorbing reactive power, and it helps reduce oscillations of power A capacitive element (leading PF) first generates and then absorbs the same amount of electric energy. Such a pattern is considered generating reactive power, and it helps maintain voltage For a real-world power system with long-distance transmission, low voltage at receiving ends is usually a big reliability concern. Shunt/series capacitors can be added as sources of reactive power or var 11

12 Complex Power def S = P + jq = V I cosθ + jv I sinθ = V I θ = V I θ θ = VI v i * = S θ = P + Q θ Reference S is the apparent power θ=tan -1 (Q/P) When θ>0 (θ i <θ v, i.e. lagging PF), Q>0 i.e. absorbing Q (inductive load) When θ<0 (θ i >θ v, i.e. leading PF), Q<0 i.e. generating Q (capacitive load) 1

13 PF=cosθ=P/ S (lagging/leading is told by +/- sign of Q) P= S cosθ= S PF = Q= = S sin θ S 1 PF, if lagging S sin θ S 1 PF, if leading If the load impedance is Z, i.e.v=zi, then S = VI * = ZII * = Z I = R I +jx I = P + jq P=R I Q=X I Z R X The load impedance angle θ is also called power angle (ϕ in some literature) Apparent power S I indicates heating and is used as a rating unit of power equipment 13

14 Some useful equations: * * VV V * * * S = VI = = = V Y Z Z Z = V S * Y = S V * If Z is purely resistive P = V R If Z is purely reactive Q V V V = = or X ωl 1/ ωc 14

15 Complex Power Balance S = VI = V[ I + I + I ] = VI + VI + VI = S + S + S * * * * P + jq = P + jq + P + jq + P + jq Due to the conservation of energy, the total complex power delivered to the loads in parallel is the sum of complex powers delivered to each (by KCL). In other words, a balance must be maintained all the time for both real power and reactive power 15

16 Example. V = V, Z = 60 + j0 Ω, Z = 6 + j1 Ω, Z = 30 j30ω 3 1 Find the power absorbed by each load and the total complex power I I I 1 3 V = = = = 0 + j0 A Z j0 60 V (1 j) = = = = = 40 j80 A Z 6 + j1 1+ j 5 V (1 + j) = = = = = 0 + j0 A Z 30 j30 1 j 3 S = VI = (0 j0) = 4, 000 W+ j0 var * 1 1 * * 3 3 S = VI = (40 + j80) = 48, 000 W+ j96, 000 var S = VI = (0 j0) = 4, 000 W j4, 000 var S = S1 + S + S3 = 96,000 W+ j7,000 var 16

17 Other approaches I = I + I + I = (0 + j0) + (40 j80) + (0 + j0) 1 3 = 80 j60 = A S (100 0 )( ) 10, VA * = VI = = = 96,000 W + j7,000 var S S S V (100) = = = 4,000 W + j0 var 60 1 * Z1 V (100) = = = 48,000 W + j96,000 var 6 j1 * Z V (100) = = = 4,000 W j4,000 var 30 + j30 3 * Z3 17

18 Theorem of Conservation of Complex Power For a network supplied by independent sources all at the same frequency (all voltages and currents are assumed to be sinusoids), the sum of the complex power supplied by the independent sources equals the sum of the complex power received by all the other branches of the network For a single source with elements in series or parallel, it is proved by Kirchhoff s voltage or current law (KVL/KCL) For a general case, it is proved by Tellegen s theorem. Application of the theorem: a part of the network can be replaced by an equivalent independent source 18

19 Some examples on Bergen and Vittal s book 19

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22 Homework # Read through Saadat s Chapter.1~.4 ECE51:.1~.6 ECE41:.1,.3,.4,.5 Due date: 9/11 (Wednesday) submitted in class or by

23 Power Factor Correction P= V I cosθ= S PF When PF<1, current I needs to increase by 1/PF to deliver the same P as that with PF=1. Thus, the major loads of the system are expected to have close to unity PFs. That is not a problem for residential and small commercial customers. However, industrial loads are inductive with low lagging PFs, so capacitors may be installed to improve PFs 3

24 Example.3 P+jQ (a) (b) Find P, Q, I and PF at the source Find capacitance C of the shunt capacitor to improve the overall PF to 0.8 lagging C Solution: (a) I I = = 0 A = = 4 j8 A 10 + j0 S = VI = 00 0 ( j0) = 400 W+ j0 var * 1 1 S = VI = 00 0 (4 + j8) = 800 W + j1600 var * S = P + jq = j1600 = VA * S I = = = A * V 00 0 PF = cos(53.13 ) = 0.6 lagging 4 (b) New power angle θ 1 = cos (0.8) = Q = P tanθ = 100 tan(36.87 ) = 900 var Q Z c c = = 700 var V (00) = = = j57.14 Ω S j700 * c 6 10 C = = 46.4 μf π (60)(57.14)

25 New apparent power and current S = j900 = * S I = = = * V Learn Example.4 (high-voltage) High-voltage capacitor bank (150kV - 75MVAR) (Source: wikipedia.org) 5

26 Complex Power Flow V1 = V1 δ1 V = V δ P 1 +jq 1 P 1 +jq 1 I 1 = V δ V δ 1 1 Z γ V1 V = δ1 γ δ γ Z Z * V1 V S1 = VI 1 1 = V1 δ 1[ 1 ] Z γ δ Z γ δ V1 V1 V = γ γ + δ1 δ Z Z P V V V = cosγ cos( γ + δ δ ) Z Z Q V V V = sin γ sin( γ + δ δ ) Z Z If R=0, i.e. Z =X and γ=90 o P V V = sin( δ δ ) X V1 Q1 = [ V1 V cos( δ1 δ )] X 6

27 P V V = sin( δ δ ) X V1 Q1 = [ V1 V cos( δ1 δ )] X Since R=0, there are no transmission line losses, and P 1 =-P 1 In a normal power systems, V i 100% and δ 1 -δ is small (e.g. 5~30 o ). P 1 sin(δ 1 -δ ) δ 1 -δ Real power flow is governed mainly by the angle difference of the terminal voltages. For example, if V 1 leads V, i.e. δ 1 -δ >0, the real power flows from node 1 to node. Theoretical maximum power transfer (i.e. static transmission capacity): V V P1 max = sin 90 = X V V X 1 1 Q 1 V 1 - V cos(δ 1 -δ ) V 1 - V Reactive power flow is determined by the magnitude difference of terminal voltages 7

28 Example.5 Determine the real and reactive powers supplied or received by each source and the power loss in the line S 1 =P 1 +jq 1 S 1 =P 1 +jq 1 I 1 V = 10 5 V, V = V, Z = 1 + j7 Ω 1 I I = = A 1+ j = = A 1+ j P S = VI = = 97.5 W + j363.3 var * S = VI = = W j94.5 var * S = S1 + S1 = 9.8 W + j68.8 var = P + jq PL QL L L L = RI1 = (1)(3.135) = 9.8 W = X I1 = (7)(3.135) = 68.8 var P, Watts P L P Source #1 Voltage Phase Angle, δ 1

29 Balanced Three-Phase Circuits A balanced source: three sinusoidal voltages are generated having the same amplitude but displaced in phase by 10 o Instantaneous power delivered to the external loads is constant Positive phase sequence Negative phase sequence 9

30 Balanced Y-connected Loads EAn = Ep α A EBn = Ep α A 10 ECn = Ep α A 40 VAn = EAn ZG Ia Van = VAn ZLIa Let V an be the reference: Phase voltages (line-to-neutral voltages): V an = V 0 V V 10 p bn = V V 40 p cn = p Line voltages (line-to-line voltages): V = V V = V ( ) = 3 V 30 ab an bn p p V = V V = V ( ) = 3 V 90 bc bn cn p p V = V V = V ( ) = 3 V 150 ca cn an p p 30 V = 3V L V bc p

31 I I a I b c Van Van 0 = = = Ip θ Z Z θ p p Vbn = = Ip 10 θ Z p Vcn = = Ip 40 θ Z p I L = I p On the neutral line (return line) I n =I a +I b +I c =0 The neutral line may not actually exist (one line per phase) The balanced 3-phase power system problems can be solved on a per-phase basis, e.g. for Phase A only. The other two phases carry identical currents except for phase shifts. 31

32 Balanced -connected Loads V L = V p Let I ab be the reference: I I ab bc I ca = I 0 p = I 10 p = I 40 p I = I I = I ( ) = 3 I 30 a ab ca p p I = I I = I ( ) = 3 I 150 b bc ab p p I = I I = I ( ) = 3 I 90 c ca bc p p I = 3 L I p 3

33 -Y Transformation Idea: compare I a =f (V an, Z ) and I a =f Y (V an, Z Y ) Z Z n Find I a =f (V an, Z ): Z I a V V V + V = + = Z Z Z ab ac ab ac I a = 3V an Z V ca Find I a =f Y (V an, Z ): I a = V Z an Y V + V = 3 V V 30 ab ac an an = 3 V ( 3/ + j/+ 3/ j/) = 3V an an Z Y = Z 3 33

34 Balanced Three-Phase Power v = V cos( ωt+ θ ) an p v v = V cos( ωt+ θ 10 ) bn p v v = V cos( ωt+ θ 40 ) cn p v For balanced loads i = I cos( ωt+ θ ) an p i i = I cos( ωt+ θ 10 ) bn p i i = I cos( ωt+ θ 40 ) cn p i Total instantaneous power: p = 3 vani + a vbni + φ b vcnic = V I cos( ωt+ θ )cos( ωt+ θ ) p p v i + V I cos( ωt+ θ 10 ) cos( ωt+ θ 10 ) p p v i + V I cos( ωt+ θ 40 ) cos( ωt+ θ 40 ) p p v i p3 φ = Vp Ip [cos( θv θi) + cos( ωt+ θv + θi)] + V I [cos( θ θ ) + cos(ωt+ θ + θ 40 )] p p v i v i + V I [cos( θ θ ) + cos(ωt+ θ + θ 480 )] p p v i v i Zero Total instantaneous power is constant: p3 = P3 = 3 Vp Ip cos φ φ θ Where is reactive power? 34

35 Extend the concept of complex power to 3-phase systems Q3 φ = 3 Vp Ip sinθ S = P + jq = V I * 3φ 3φ 3φ 3 p p Expressed in terms of line voltage V L and line current I L Y-connection: V = V / 3 Ip = IL -connection: p V p L = V I = I / 3 L p L P3 φ = 3 VL IL cosθ Q3 φ = 3 VL IL cosθ θ is the angle between the phase voltage and the phase current for the same phase (e.g. Phase A) 35

36 Example.7 (a) (b) (c) (d) The current, real power and reactive powers drawn from the supply The line voltage at the combined loads The current per phase in each load The total real and reactive powers in each load and the line 36

37 V =110-j0 37

38 Homework #3 Read through Saadat s Chapter.5~.1 ECE51:.7~.10,.14~.16 ECE41:.7,.8,.10,.15,.16 Due date: 9/0 (Friday) submitted in class or by 38